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wlad13 [49]
2 years ago
13

Astronauts brought back 500 lb of rock samples from the moon. how many kilograms did they bring back? 1 kg = 2.20 lb 227 kg 227

kg 498 kg 498 kg 500 kg 500 kg 1,100 kg
Physics
1 answer:
Aleksandr [31]2 years ago
7 0

Astronauts brought back 227kg of rock samples from the moon.

<h3>Who is an astronaut?</h3>

A person who has been trained, outfitted, and sent into space by a human spaceflight program to serve as a commander or crew member is known as an astronaut (from the Ancient Greek words (astron), meaning "star," and (nautes), meaning "sailor"). The word is occasionally used to refer to anyone who flies into space, including scientists, politicians, journalists, and tourists, despite the fact that it is typically only designated for professional space travelers.

600 astronauts have made space flights since 1961. Up until 2002, only governments—either the military or civilian space agencies—sponsored and educated astronauts. A new class of astronaut was introduced in 2004 with the suborbital flight of the privately funded SpaceShipOne: the commercial astronaut.

Explanation:

1kg=2.20lb

500lb=(1/2.20*500)kg

500lb=227kg

To learn more about astronaut , visit:

brainly.com/question/11244838

#SPJ4

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Natalija [7]
No, Their arms and back legs are too short, but the mesosaurus was also a coastal animal! It's very unlikely.
3 0
3 years ago
A toy cannon uses a spring to project a 5.35-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force
Elenna [48]

Answer:

a) the velocity is v=1.385 m/s

b) the ball has its maximum speed at 4.68 cm away from its compressed position

c)  the maximum speed is 1.78 m/s

Explanation:

if we do an energy balance over the ball, the potencial energy given by the compressed spring is converted into kinetic energy and loss of energy due to friction, therefore

we can formulate this considering that the work of the friction force is equal to to the energy loss of the ball

W fr = - ΔE = - ΔU - ΔK = Ui - Uf + Kf - Ki

therefore

Ui + Ki = Uf + Kf + W fr  

where U represents potencial energy of the compressed spring , K is the kinetic energy W fr is the work done by the friction force. i represents inicial state, and f final state.

since

U= 1/2 k x² , K= 1/2 m v²  , W fr = F*L

X= compression length , L= horizontal distance covered

therefore

Ui + Ki = Uf + Kf + W fr

1/2 k xi² + 1/2 m vi² = 1/2 k x² + 1/2 m vf² + F*L

a) choosing our inicial state as the compressed state , the initial kinetic energy is Ki=0 and in the final state the ball is no longer pushed by the spring thus Uf=0

1/2 k X² + 0 = 0 + 1/2 m v² + F*L

1/2 m v² = 1/2 k X² - F*L

v = √[(k/m)x² -(2F/m)*L] = √[(8.07N/m/5.35*10^-3 Kg)*(-0.0508m)² -(2*0.033N/5.35*10^-3 Kg)*(0.16 m)] = 1.385 m/s

b) in any point x , and since L= d-(X-x) , d = distance where is no pushed by the spring.

1/2 k X² + 0 = 1/2 k x² + 1/2 m v² + F*[d-(X+x)]

1/2 m v² =1/2 k X²-1/2 k x² - F*[d-(X-x)] = (1/2 k X²+ F*X) - 1/2k x² - F*x + F*d

taking the derivative

dKf/dx = -kx - F = 0 → x = -F/k = -0.033N/8.07 N/m = -4.089*10^-3 m = -0.4cm

at x m = -0.4 cm the velocity is maximum

therefore is 5.08 cm-0.4 cm=4.68 cm away from the compressed position

c) the maximum speed is

1/2 m v max² = (1/2 k X²+ F*X) - 1/2k x m² - F*(x m) + 0

v =√[ (k/m) (X²-xm²) + (2F/m)(X-xm) ] = √[(8.07N/m/5.35*10^-3 Kg)*[(-0.0508m)² - (-0.004m)²] + (2*0.033N/5.35*10^-3 Kg)*(-0.0508m-(-0.004m)] = 1.78 m/s

4 0
3 years ago
Ten seconds after an electric fan is turned on, the fan rotates at 300 rev/min. its average angular acceleration is
Tcecarenko [31]
Check the attached file for the answer.

6 0
3 years ago
A force has both______ and _______. Question 1 options: Speed and impact Magnitude and direction Motion and speed Direction and
Travka [436]

Answer:

magnitude

and direction

Explanation:

5 0
2 years ago
Resistors 1 and 2− R1 = 50 Ω , R2 = 90 Ω − are connected in series to a 6.0-V battery. Part APart complete What is the potential
kondor19780726 [428]

Answer:

Part A: The voltage across resistor R1 is approximately \rm 2.1 \; V.

Part B: When the value of resistor R1 decreases, the current in this circuit will increase.

Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.

Explanation:

<h3>Part A</h3>

Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of R_1 + R_2 = 50 + 90 = 140\; \Omega. The voltage across the two resistor, combined, is equal to \rm 6\; V. Hence by Ohm's Law, the current through the circuit will be equal to \rm \dfrac{6\; V}{140\; \Omega} = \dfrac{3}{70}\; A.

These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to \rm \dfrac{3}{70}\; A. Apply Ohm's Law (again) to find the voltage across R1:

V = I \cdot R = \dfrac{3}{70} \times 50 \approx \rm 2.1\; V.

<h3>Part B</h3>

Since the equivalent resistance is equal to R_1 + R_2, when the value of R_1 decreases, the equivalent resistance will also decrease. By Ohm's Law, I = \dfrac{V}{R}. When the value of the denominator ( decreases, the value of the quotient, I the current through the circuit, will increase.

<h3>Part C</h3>

Keep in mind that if two resistors are connected in series,

I(R_1) = I(\text{Circuit}) = I(R_2).

The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)

Again, since the two resistors are connected in series,

V(R_1) + V(R_2) = V(\text{Circuit}) = \rm 6 \; V,

when the voltage across R2 increases, the voltage across R1 will decrease.

4 0
3 years ago
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